# Question #3b185

Nov 17, 2016

See below.

#### Explanation:

Calling $f \left(x\right) = x - e {\log}_{e} x$ it's minimum is located at

$\frac{\mathrm{df}}{\mathrm{dx}} = 1 - \frac{e}{x} = 0$

or at $x = e$

At this point we have

$f \left(e\right) = 0$

To qualify this stationary point we compute

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{e}{x} ^ 2$ and for $x = e$ we have

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = {e}^{- 1} > 0$ so $x = e$ represents a minimum of $f \left(x\right)$

The conclusion is:

$e {\log}_{e} x \le x$ for $x > 0$ and also $e {\log}_{e} x$ and $x$ osculate at $x = e$.

Note that for $x < 0$, ${\log}_{e} x$ is not defined as a real function of a real variable.

Attached a plot with $e {\log}_{e} x$ (blue) and $x$ (red)